Let a = 2i - j - k, b = i + 3j - k and c = 2i + j + 3k. Let v be the vector in the plane of the vect

Q: Why did we set $\lambda = 1$ in $\vec{v} = \lambda \vec{a} + \mu \vec{b}$? Isn't that arbitrary?

That is a brilliant observation! In this specific problem, we are looking for a vector $\vec{v}$ that satisfies the condition. If we chose a different $\lambda$, we would simply be scaling the entire vector $\vec{v}$, which would change the projection length. By setting $\lambda = 1$, we are essentially fixing the 'scale' of the vector to find the specific magnitude required. If you kept $\lambda$ as a variable, you would find that the final magnitude $|\vec{v}|$ would be a function of $\lambda$, but the problem implies a unique solution, suggesting the setup is consistent with $\lambda=1$.

Q: Why does the projection formula use the absolute value $|\vec{v} \cdot \vec{c}|$?

Because projection length is a distance, and distance must always be non-negative! The dot product $\vec{v} \cdot \vec{c}$ can be negative if the angle between the vectors is obtuse (greater than 90 degrees). The absolute value ensures we are talking about the magnitude of the projection, not the signed scalar component.

Q: Is there a faster way than using the projection formula?

You could use the cross product to find the normal to the plane, but that is often more computationally expensive. The beauty of this problem lies in the dot product. When you see 'projection' and 'plane', always look for ways to use the dot product properties first. It is the most direct path to the solution.

Question

Let $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}, \vec{b} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to

Options:

(A) 13

(B) $\frac{\sqrt{35}}{2}$

(C) $\frac{\sqrt{21}}{2}$

(D) 7

Sigma Percentile · Accepted Answer

### The Geometry of Space: A Vector Odyssey

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are navigating the elegant architecture of three-dimensional space.

We have three vectors: $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$, and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$. We are tasked with finding a mysterious vector $\vec{v}$ that resides in the plane defined by $\vec{a}$ and $\vec{b}$, constrained by its projection onto $\vec{c}$.

### Phase 1: The Hidden Orthogonality

Before we write a single equation, let us observe the vectors. In JEE Advanced, problems are rarely random; there is almost always a hidden symmetry. Let us test the relationship between $\vec{a}$ and $\vec{b}$ using the dot product:

$$\vec{a} \cdot \vec{b} = (2)(1) + (-1)(3) + (-1)(-1) = 2 - 3 + 1 = 0$$

Stop and breathe. The dot product is zero. This is not a coincidence; it is a gift, telling us that $\vec{a}$ and $\vec{b}$ are perfectly perpendicular.

Now, let us check $\vec{a}$ and $\vec{c}$:

$$\vec{a} \cdot \vec{c} = (2)(2) + (-1)(1) + (-1)(3) = 4 - 1 - 3 = 0$$

Again, zero! This means $\vec{a}$ is orthogonal to both $\vec{b}$ and $\vec{c}$. We have just simplified our geometric landscape significantly, as $\vec{a}$ acts as a clean, independent axis.

### Phase 2: Constructing the Vector $\vec{v}$

Since $\vec{v}$ lies in the plane of $\vec{a}$ and $\vec{b}$, we can define it as a linear combination of these two vectors:

$$\vec{v} = \lambda \vec{a} + \mu \vec{b}$$

For the sake of finding the specific vector required by the problem, we set $\lambda = 1$. Thus, our vector becomes $\vec{v} = \vec{a} + \mu \vec{b}$.

Imagine this vector $\vec{v}$ as a path on a flat surface defined by the vectors $\vec{a}$ and $\vec{b}$. We need to find the value of $\mu$ that satisfies the projection constraint.

### Phase 3: The Projection Constraint

The problem states that the length of the projection of $\vec{v}$ on $\vec{c}$ is $\frac{1}{\sqrt{14}}$. The formula for the projection of a vector $\vec{v}$ onto $\vec{c}$ is:

$$	ext{Projection} = \frac{|\vec{v} \cdot \vec{c}|}{|\vec{c}|}$$

Let us expand the numerator, $\vec{v} \cdot \vec{c}$, using our definition of $\vec{v}$:

$$\vec{v} \cdot \vec{c} = (\vec{a} + \mu \vec{b}) \cdot \vec{c} = \vec{a} \cdot \vec{c} + \mu (\vec{b} \cdot \vec{c})$$

Remember our earlier discovery that $\vec{a} \cdot \vec{c} = 0$. The first term vanishes into thin air, leaving us with $\mu (\vec{b} \cdot \vec{c})$. Let us calculate $\vec{b} \cdot \vec{c}$:

$$\vec{b} \cdot \vec{c} = (1)(2) + (3)(1) + (-1)(3) = 2 + 3 - 3 = 2$$

So, $\vec{v} \cdot \vec{c} = 2\mu$. Now, we calculate the magnitude of $\vec{c}$:

$$|\vec{c}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$$

### Phase 4: The Final Calculation

We equate our findings to the given projection length:

$$\frac{|2\mu|}{\sqrt{14}} = \frac{1}{\sqrt{14}}$$

The $\sqrt{14}$ terms cancel out beautifully, leaving us with $|2\mu| = 1$, which means $\mu = \pm \frac{1}{2}$.

Finally, we need the magnitude of $\vec{v}$. Since $\vec{a}$ and $\vec{b}$ are perpendicular, we can use the Pythagorean theorem for vectors:

$$|\vec{v}|^2 = |\vec{a}|^2 + \mu^2 |\vec{b}|^2$$

Calculating the magnitudes:

$$|\vec{a}|^2 = 2^2 + (-1)^2 + (-1)^2 = 6$$
$$|\vec{b}|^2 = 1^2 + 3^2 + (-1)^2 = 11$$

Substituting $\mu^2 = (\frac{1}{2})^2 = \frac{1}{4}$:

$$|\vec{v}|^2 = 6 + \frac{1}{4}(11) = 6 + \frac{11}{4} = \frac{24 + 11}{4} = \frac{35}{4}$$

Taking the square root, we arrive at the final answer:

**$$|\vec{v}| = \frac{\sqrt{35}}{2}$$**

And there it is. Through the power of orthogonality and the elegance of the dot product, we have navigated the problem. Keep this mindset—always look for the simplest path, and the math will reveal itself.

Select Answer:

Visualized Solution

Define the Given Vectors

Check Orthogonality of $a$ and $b$

Check Orthogonality of $a$ and $c$

Express $v$ in the Plane of $a$ and $b$

Calculate Dot Product $v \cdot c$

Calculate Magnitude of $c$

Apply Projection Formula

Calculate Magnitude of $v$

Final Answer

The Sigma Insight: Scalar (Dot) Product

The Geometry of Space

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three vectors. A vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{3}$ , is given by

Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + 2 \hat{j} - \hat{k}$ and $c = \hat{i} + \hat{j} - 2 \hat{k}$ be three vectors. A vector in the plane of $b$ and $c$ , whose projection on $a$ is of magnitude $2/3$ , is:

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} + \hat{j} - \hat{k}$ . A vector in the plane of $a$ and $b$ whose projection on $c$ is $\frac{1}{3}$ , is

Let $u, v, w$ be such that $∣ u ∣ = 1, ∣ v ∣ = 2, ∣ w ∣ = 3$ . If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v, w$ are perpendicular to each other then $∣ u - v + w ∣$ equals

Let $a = \hat{i} - \hat{j} + 2 \hat{k}$ and $b$ be a vector such that $a \times b = 2 \hat{i} - \hat{k}$ and $a \cdot b = 3$ . Then the projection of $b$ on the vector $a - b$ is :-

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

Let $x$ be a vector in the plane containing vectors $a = 2 \hat{i} - \hat{j} + \hat{k}$ and $b = \hat{i} + 2 \hat{j} - \hat{k}$ . If the vector $x$ is perpendicular to $(3 \hat{i} + 2 \hat{j} - \hat{k})$ and its projection on $a$ is $\frac{17 6}{2}$ , then the value of $∣ x ∣^{2}$ is equal to

Let $a = 2 \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + 2 \hat{j} + \hat{k}$ be two vectors. Consider a vector $c = α a + β b, α, β \in R$ . If the projection of $c$ on the vector $(a + b)$ is $32$ , then the minimum value of $(c - (a \times b)) \cdot c$ equals